Integral Evaluation: Solving \int^3_0 \sqrt{t^4+t^2} dt

[roam]

Homework Statement

This is part of a larger problem about finding the distance traveled by the particle over the interval 0≤t≤3. I need to solve the integral

$\int^3_0 \sqrt{t^4+t^2} \ dt$

The Attempt at a Solution

So, is it correct to rewrite $\sqrt{t^4+t^2}$ as $t \sqrt{t^2+1}$ and then use integration by parts?

I'm confused because when I use Wolfarm online integrator to evaluate

$\int \ \sqrt{t^4+t^2} = \frac{(t^2+1) \sqrt{t^4+x^2}}{3t}$

But when I use the other expression I get:

$\int \ t \sqrt{t^2+1} = \frac{1}{3} (t^2+1)^{3/2}$

So which one is correct?

 

[QuarkCharmer]

Both seem to be valid. You had the right idea to calculate the integral by hand. Factor out the t^2, then remove it from the radical leaving $t\sqrt{t^{2}+1}$ and use integration by parts.

 

[Mark44]

Homework Statement

This is part of a larger problem about finding the distance traveled by the particle over the interval 0≤t≤3. I need to solve the integral

$\int^3_0 \sqrt{t^4+t^2} \ dt$

The Attempt at a Solution

So, is it correct to rewrite $\sqrt{t^4+t^2}$ as $t \sqrt{t^2+1}$ and then use integration by parts?

Yes, it can be rewritten this way. I would use an ordinary substitution, though, not integration by parts.

I'm confused because when I use Wolfarm online integrator to evaluate

$\int \ \sqrt{t^4+t^2} = \frac{(t^2+1) \sqrt{t^4+x^2}}{3t}$

There's a typo above - you shouldn't have x in the antiderivative.

But when I use the other expression I get:

$\int \ t \sqrt{t^2+1} = \frac{1}{3} (t^2+1)^{3/2}$

So which one is correct?

 

[roam]

Yes, it can be rewritten this way. I would use an ordinary substitution, though, not integration by parts.
There's a typo above - you shouldn't have x in the antiderivative.

That's a typo, I meant to say: $\frac{(t^2+1)\sqrt{t^4+t^2}}{3t}$

So, my question was: shouldn't the final intrgral be unique regardless of how I simplify the integrand? Because I got two clearly different answers. Which method was correct?

 

[DryRun]

The final integral will depend on your integration technique, but should be unique after simplification.

The second one is correct as it's in its most simplified form. Wolfram's format is computer-generated, so although different, it's equivalent.

 

[Mark44]

If you use two techniques to evaluate an indefinite integral (no limits of integration), you won't necessarily get the same things, but they can differ by at most a constant.

If you use two techniques to evaluate a definite integral, they have to produce the same result (unless you make a mistake).